Question

Consider the four variables A, B, C and D and a function Z of these variables, Z=15A\(^2\)-3B\(^4\)+C+0.5D It is given that A, B, C and D must be non-negative integers and that all of the following relationships must hold:
i) \(2A+B\leq2\)
ii) \(4A+2B+C\leq12\)
iii) \(3A+4B+D\leq15\)
If Z needs to be maximised, then what value must D take?

• A. 15
• B. 12
• C. 0
• D. 10
• E. 5

Answer 1

The Correct Option is B

To solve the problem of maximizing the function \(Z=15A^2-3B^4+C+0.5D\) under the given constraints, we'll analyze each condition and determine feasible values for \(A\), \(B\), \(C\), and \(D\).

Let's examine the constraints:

• \(2A+B\leq2\)
• \(4A+2B+C\leq12\)
• \(3A+4B+D\leq15\)

Firstly, possible values for \(A\) and \(B\) based on constraint (i) are:

• If \(A=0\), then \(B\leq2\) implying \(B\) can be 0, 1, or 2.
• If \(A=1\), then \(B\leq0\) implying \(B=0\).
• \(A=2\) isn't permissible as \(2A+B\) would exceed 2.

For each pair of \((A,B)\), we check values of \(C\) and \(D:\)

• For \((A=0,B=0)\):
  \(4A+2B+C\leq12\) implies \(C\leq12\).
  \(3A+4B+D\leq15\) implies \(D\leq15\).
  \(Z=C+0.5D\).
• For \((A=0,B=1)\):
  \(4A+2B+C\leq12\) implies \(C\leq10\).
  \(3A+4B+D\leq15\) implies \(D\leq11\).
  \(Z=C+0.5D-3\).
• For \((A=0,B=2)\):
  \(4A+2B+C\leq12\) implies \(C\leq8\).
  \(3A+4B+D\leq15\) implies \(D\leq7\).
  \(Z=C+0.5D-48\) (as \(3\times16=48\)).
• For \((A=1,B=0)\):
  \(4A+2B+C\leq12\) implies \(C\leq8\).
  \(3A+4B+D\leq15\) implies \(D\leq12\).
  \(Z=15+C+0.5D\).

To maximize \(Z\), we substitute the maximum allowable values of \(C\) and \(D\) based on their constraints:

• For \((A=1,B=0)\) with \(C=8\), \(D=12\):
  \(Z=15+8+0.5\times12=29\)

Testing other feasible divisions will not yield a higher \(Z\) value than 29 based on calculation for remaining possibilities.

Thus, the value \(D\) must take to maximize \(Z\) is 12.